Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $q \neq 0$. $p = \dfrac{-q + 8}{7q + 28} \div \dfrac{q^2 - 3q - 40}{10q + 50} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $p = \dfrac{-q + 8}{7q + 28} \times \dfrac{10q + 50}{q^2 - 3q - 40} $ First factor the quadratic. $p = \dfrac{-q + 8}{7q + 28} \times \dfrac{10q + 50}{(q - 8)(q + 5)} $ Then factor out any other terms. $p = \dfrac{-(q - 8)}{7(q + 4)} \times \dfrac{10(q + 5)}{(q - 8)(q + 5)} $ Then multiply the two numerators and multiply the two denominators. $p = \dfrac{ -(q - 8) \times 10(q + 5) } { 7(q + 4) \times (q - 8)(q + 5) } $ $p = \dfrac{ -10(q - 8)(q + 5)}{ 7(q + 4)(q - 8)(q + 5)} $ Notice that $(q + 5)$ and $(q - 8)$ appear in both the numerator and denominator so we can cancel them. $p = \dfrac{ -10\cancel{(q - 8)}(q + 5)}{ 7(q + 4)\cancel{(q - 8)}(q + 5)} $ We are dividing by $q - 8$ , so $q - 8 \neq 0$ Therefore, $q \neq 8$ $p = \dfrac{ -10\cancel{(q - 8)}\cancel{(q + 5)}}{ 7(q + 4)\cancel{(q - 8)}\cancel{(q + 5)}} $ We are dividing by $q + 5$ , so $q + 5 \neq 0$ Therefore, $q \neq -5$ $p = \dfrac{-10}{7(q + 4)} ; \space q \neq 8 ; \space q \neq -5 $